Kinetic Theory

JEE Main level, single correct, medium

33 of 168

Question 33 of 168

The root mean square speed of molecules of nitrogen gas at 27∘C27^{\circ}C is approximately: (Given mass of a nitrogen molecule =4.6×10−26kg=4.6\times {10}^{-26}kg and take Boltzmann constant kB=1.4×10−23JK−1{k}_{B}=1.4\times {10}^{-23}J{K}^{-1} )

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Answer: (D) 523ms−1523m{s}^{-1}

The RMS value of a gas is given by
vrms=3RTM=3kBTm{v}_{rms}=\sqrt{\dfrac{3RT}{M}} =\sqrt{\dfrac{3{k}_{B}T}{m}}
Substituting the given values,
vrms=3×1.4×10−23×3004.6×10−26{v}_{rms}=\sqrt{\dfrac{3\times 1.4\times {10}^{-23}\times 300}{4.6\times {10}^{-26}}}
≈523ms−1\approx 523m{s}^{-1}

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